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uranmaximum [27]

3 years ago

11

Wo gravitational forces act on a particle, in perpendicular directions. to find the net force, can we add the magnitudes of thos

e two forces?

1 answer:

son4ous [18]3 years ago

3 0

No, you can not.

$f=mxa$

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Chris threw a basketball a distance of 27.5 m to score and win his

salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

$R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}$

<u>v₀ = 16.55 m/s</u>

8 0

3 years ago

A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5m/s2 for 4s. After t

Ber [7]

Answer:

+7.0 m/s

Explanation:

Let's take rightward as positive direction.

So in this problem we have:

a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)

t = 4 s time interval

v = -3.0 m/s is the final velocity (negative because it is leftward)

We can use the following equation:

v = u + at

Where u is the initial velocity

We want to find u, so if we rearrange the equation we find:

$u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s$

and the positive sign means the initial direction was rightward.

6 0

4 years ago

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

7 0

3 years ago

A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l

mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be $\who_w = 1000 kg/m^3$

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

$F_s + F_b = W$

Where $F_s = kx$ N is the spring force, $F_b = W_w = m_wg = \rho_w V_s g$ is the buoyancy force, which equals to the weight $W_w$ of the water displaced by the submerged portion of the cylinder, which is the product of water density $\rho_w$ , submerged volume $V_s$ and gravitational constant g. W = mg is the weight of the metal cylinder.

$kx + \rho_w V_s g = mg$

The submerged volume would be the product of cross-section area and the submerged length x

$V_s = Ax = \pi(d/2)^2x$

Plug that into our force equation and we have

$kx + \rho_w \pi(d/2)^2x g = mg$

$x(k + \rho_w g \pi d^2/4) = mg$

$x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m$

6 0

3 years ago

A crate slides down a ramp that makes a 20∘ angle with the ground. To keep the crate moving at a steady speed, Paige pushes back

VMariaS [17]

Answer:

Hence, work done= 287.54 J

Explanation:

Given data:

angle of ramp with the ground θ =20°

force applied = 76 N

work done on the crate to slide down 4 m down the ramp

W= F×d cosθ ( only the cos component of the force will slide the crate down)

W= 76×4×cos20= 287.54 J

4 0

4 years ago